What is the limit of # (1+2x)^(1/x)# as x approaches infinity?

Answer 1

1

#lim_(x to oo) (1+2x)^(1/x)#

We can use exponentials and logs here:

#= lim_(x to oo) e^( ln(1+2x)^(1/x))#
#= lim_(x to oo) e^( ( ln(1+2x))/x)#
Because #e# is a continuous function in #(-oo, oo)#, we can also say this:
#= e^( lim_(x to oo) color(red)(( ln(1+2x))/x))#

....and we can focus on the term in red.

We know straightaway that, because the log will grow more slowly than the #x# term, then the term in red will go to zero; and so the limit is #1#.
We can use L'Hopital's Rule to see this out as this is in #oo/oo# indeterminate form. One application of L'H takes us here:
# ln(1+2x)/x to (2/(1+2x))/1 = 2/(1+ 2x)#

And so our limit is:

#= e^( lim_(x to oo) ( 2/(1+2x)))#
#= e^( lim_(x to oo) 0) = 1#
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Answer 2

The limit of (1+2x)^(1/x) as x approaches infinity is e^2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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